Q. A tower subtends an angle of 30$^\circ$ at a point on the same level as the foot of the tower. At a second point h metres above the first, the depression of the foot of the tower is 60$^\circ$. The horizontal distance of the tower from the point is.
Solution:
Let PQ = $x$ metres denote the tower so that $\Delta PAQ = 30^\circ$. Let AB = $h$ metres. $\therefore \, \Delta BQA = 60^\circ$
Now, $\frac{h}{AQ} $ = tan $60^\circ = \sqrt{3}$
$\therefore $ AQ = $\frac{h}{\sqrt{3}} = h$ cot $60^\circ$
